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Understanding mean and variance is essential in exploring the normal distribution. Let's start with the mean. In this exercise, each candy's weight is normally distributed with a mean of 0.1 ounce. The mean of a distribution tells us the average value. So, when we have 16 candies, we simply multiply this mean by 16 to get the average weight of all candies together, which sums up to 1.6 ounces.
Variance, on the other hand, measures the spread or dispersion of the data. It tells us how much the individual weights differ from the mean. For one candy, the variance is the square of the standard deviation: \(0.01^2 = 0.0001\). When considering 16 independent candies, the variance of their total weight is \(16 \times 0.0001 = 0.0016\) ounces squared. This larger variance accounts for more fluctuations as the number of candies increases.
The standard deviation, which is the square root of the variance, shows us the average difference from the mean. Here, it calculates to \(\sqrt{0.0016} = 0.04\) ounces, helping us understand the average spread relative to the mean.

In probability calculations involving the normal distribution, symmetry about the mean plays a crucial role. Given that the mean weight of our package is 1.6 ounces, we want to determine the probability that the package's net weight is less than this mean.
Thanks to the properties of the normal distribution, this probability is straightforward. Since our mean value, 1.6 ounces, lies exactly at the center, the probability that the package weight is less than 1.6 ounces is 0.5. This is because half of the distribution lies on each side of the mean.
Analyzing probabilities with a normal distribution can be handy, especially when quantities naturally cluster around a central mean with some variability. Here, knowing that the distribution is symmetric allows us to easily find the probability that falls on either side of the mean without complex calculations.

The Z-score is a statistical measure that tells us how many standard deviations an element is from the mean. It helps standardize values for comparison within the context of a normal distribution.
In this exercise, you want to find the weight that the mean of the candies will exceed with 99% probability. First, identify the Z-score that matches a cumulative probability of 0.99. Referring to Z-tables, this Z-score is approximately 2.33.
Once the Z-score is known, you can use it to find the corresponding weight value. Using the formula \(X = \mu + Z \sigma\), where \(\mu\) is the mean (1.6 ounces), \(Z\) is the Z-score (2.33), and \(\sigma\) is the standard deviation (0.04 ounces), you can substitute these values in. This gives you \(X = 1.6 + 2.33 \times 0.04 = 1.6932\) ounces.
Thus, with 99% certainty, the package weight will not exceed 1.6932 ounces. Understanding how to apply the Z-score transforms probability calculations from just numbers into meaningful metrics.